Integrals • Topic 1 of 2

Indefinite Integrals & Methods of Integration

Integration is the reverse of differentiation. A function $F$ is an antiderivative (or primitive) of $f$ on an interval if $F'(x)=f(x)$ for every $x$ there. Because the derivative of a constant is zero, antiderivatives are never unique: if $F'(x)=f(x)$ then $\bigl(F(x)+C\bigr)'=f(x)$ too. The whole family is the indefinite integral, written

$$\int f(x)\,dx = F(x)+C,\qquad F'(x)=f(x).$$

Here $f(x)$ is the integrand, $x$ is the variable of integration, and $C$ is the arbitrary constant of integration — omitting it is the single most common mistake in this chapter. Integration is linear: $\int\bigl(k_1f(x)+k_2g(x)\bigr)dx = k_1\int f(x)\,dx + k_2\int g(x)\,dx$. There is no product rule or quotient rule for integration; instead we build a toolbox of methods.

Standard integrals. Every formula below is just a differentiation rule read backwards, so the fastest check on any answer is to differentiate it.

$\int f(x)\,dx$	Result	$\int f(x)\,dx$	Result
$\int x^{n}\,dx\ (n\ne-1)$	$\dfrac{x^{n+1}}{n+1}+C$	$\int \dfrac1x\,dx$	$\ln\|x\|+C$
$\int e^{x}\,dx$	$e^{x}+C$	$\int a^{x}\,dx$	$\dfrac{a^{x}}{\ln a}+C$
$\int \cos x\,dx$	$\sin x+C$	$\int \sin x\,dx$	$-\cos x+C$
$\int \sec^2x\,dx$	$\tan x+C$	$\int \csc^2x\,dx$	$-\cot x+C$
$\int \sec x\tan x\,dx$	$\sec x+C$	$\int \csc x\cot x\,dx$	$-\csc x+C$
$\int \tan x\,dx$	$\ln\|\sec x\|+C$	$\int \cot x\,dx$	$\ln\|\sin x\|+C$
$\int \sec x\,dx$	$\ln\|\sec x+\tan x\|+C$	$\int \csc x\,dx$	$\ln\|\csc x-\cot x\|+C$
$\int \dfrac{dx}{\sqrt{1-x^2}}$	$\sin^{-1}x+C$	$\int \dfrac{dx}{1+x^2}$	$\tan^{-1}x+C$

1. Integration by substitution

If the integrand contains a function $g(x)$ together with (a multiple of) its derivative $g'(x)$, substitute $u=g(x)$ so that $du=g'(x)\,dx$. The integral collapses to a standard one in $u$:

$$\int f\bigl(g(x)\bigr)g'(x)\,dx \;=\; \int f(u)\,du.$$

Two patterns worth memorising as direct consequences: $\int \dfrac{f'(x)}{f(x)}\,dx=\ln|f(x)|+C$ (top is the derivative of the bottom) and $\int \bigl[f(x)\bigr]^{n}f'(x)\,dx=\dfrac{[f(x)]^{n+1}}{n+1}+C$. Always substitute back to the original variable at the end.

2. Integration using trigonometric identities

Powers and products of $\sin$ and $\cos$ are not directly integrable, so first rewrite them with identities. Use power-reduction for even powers, $\sin^2x=\dfrac{1-\cos2x}{2}$ and $\cos^2x=\dfrac{1+\cos2x}{2}$; and product-to-sum for products of different angles:

$$2\sin A\cos B=\sin(A{+}B)+\sin(A{-}B),\quad 2\cos A\cos B=\cos(A{-}B)+\cos(A{+}B),\quad 2\sin A\sin B=\cos(A{-}B)-\cos(A{+}B).$$

For odd powers of sine or cosine, peel off one factor and convert the rest using $\sin^2x+\cos^2x=1$, then substitute. Triple-angle identities such as $\sin^3x=\dfrac{3\sin x-\sin 3x}{4}$ are handy shortcuts.

3. Integrals of special rational and irrational forms

These results, obtained once by completing the square or partial fractions, become standard tools:

$$\int\frac{dx}{x^2-a^2}=\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|+C,\qquad \int\frac{dx}{a^2-x^2}=\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right|+C,$$

$$\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\frac{x}{a}+C,\qquad \int\frac{dx}{\sqrt{a^2-x^2}}=\sin^{-1}\frac{x}{a}+C,$$

$$\int\frac{dx}{\sqrt{x^2+a^2}}=\ln\left|x+\sqrt{x^2+a^2}\right|+C,\qquad \int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right|+C.$$

For a quadratic denominator $ax^2+bx+c$, complete the square to reach one of the $x^2\pm a^2$ forms. For $\dfrac{px+q}{ax^2+bx+c}$, write the numerator as $\lambda(\text{derivative of denominator})+\mu$, splitting the integral into a $\ln$ part and a $\tan^{-1}$/$\ln$ part.

4. Integration by partial fractions

A proper rational function (degree of numerator $<$ degree of denominator) splits into a sum of simpler fractions which integrate term by term. If the integrand is improper, divide first. The decomposition depends on the denominator's factors:

Factor in denominator	Partial fraction term(s)
Non-repeated linear $(x-a)$	$\dfrac{A}{x-a}$
Repeated linear $(x-a)^2$	$\dfrac{A}{x-a}+\dfrac{B}{(x-a)^2}$
Non-repeated quadratic $(x^2+bx+c)$	$\dfrac{Ax+B}{x^2+bx+c}$

Find the unknowns by clearing denominators and either equating coefficients or substituting convenient values of $x$.

5. Integration by parts (ILATE)

For a product of two unlike functions use

$$\int u\,dv = uv-\int v\,du.$$

Choose the first function $u$ (the one to differentiate) by the ILATE order — Inverse trig, Logarithm, Algebraic, Trig, Exponential — whatever comes first becomes $u$. A function standing alone, such as $\ln x$ or $\tan^{-1}x$, is integrated by parts with $dv=dx$. A valuable special result is

$$\int e^{x}\bigl[f(x)+f'(x)\bigr]dx = e^{x}f(x)+C,$$

so any integrand of the form "$e^x$ times (a function plus its own derivative)" integrates instantly. The matching results for the two square-root irrationals are

$$\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}+C,$$

$$\int\sqrt{x^2\pm a^2}\,dx=\frac{x}{2}\sqrt{x^2\pm a^2}\pm\frac{a^2}{2}\ln\left|x+\sqrt{x^2\pm a^2}\right|+C.$$

Deeper Insight — integration is pattern recognition, not a single algorithm. Differentiation is mechanical: any elementary function has an elementary derivative you can grind out with the rules. Integration is the opposite — there is no universal recipe, and some perfectly innocent functions (like $e^{-x^2}$) have no elementary antiderivative at all. So the real skill is diagnosis: look at the integrand and ask which tool fits. Is the derivative of one part sitting next to it (substitution)? Is it a product of unlike functions (by parts)? A ratio of polynomials (partial fractions)? A trig power (an identity first)? Every method here is one differentiation rule reversed, which is why you can — and should — verify any answer by differentiating it back to the integrand.

Solved Examples

Worked Example

Evaluate $\displaystyle\int (3x^2+2x)\,dx$.

Solution

Integrate term by term using the power rule: $=3\cdot\dfrac{x^3}{3}+2\cdot\dfrac{x^2}{2}+C=x^3+x^2+C.$

Answer: $x^3+x^2+C$

Worked Example

Evaluate $\displaystyle\int 2x\,e^{x^2}\,dx$.

Solution

Put $u=x^2$, so $du=2x\,dx$. Then $\displaystyle\int e^{u}\,du=e^{u}+C=e^{x^2}+C.$

Answer: $e^{x^2}+C$

Worked Example

Evaluate $\displaystyle\int x\,e^{x}\,dx$.

Solution

By parts with $u=x,\ dv=e^x\,dx$, so $du=dx,\ v=e^x$. $\displaystyle\int xe^x\,dx=xe^x-\int e^x\,dx=xe^x-e^x+C=e^x(x-1)+C.$

Answer: $e^x(x-1)+C$

Worked Example

Evaluate $\displaystyle\int \frac{1}{x^2-1}\,dx$.

Solution

Partial fractions: $\dfrac{1}{x^2-1}=\dfrac{1}{2}\left(\dfrac{1}{x-1}-\dfrac{1}{x+1}\right)$. Integrating, $=\dfrac12\ln\left|\dfrac{x-1}{x+1}\right|+C.$

Answer: $\dfrac12\ln\left|\dfrac{x-1}{x+1}\right|+C$

Worked Example

Evaluate $\displaystyle\int \frac{2x}{x^2+1}\,dx$.

Solution

The numerator is exactly the derivative of the denominator, so this fits $\int\dfrac{f'}{f}\,dx=\ln|f|+C$. Hence $=\ln(x^2+1)+C$ (no modulus needed, since $x^2+1>0$).

Answer: $\ln(x^2+1)+C$

Worked Example

Evaluate $\displaystyle\int \tan x\,dx$.

Solution

Write $\tan x=\dfrac{\sin x}{\cos x}$. Put $u=\cos x$, $du=-\sin x\,dx$, so $\displaystyle\int\tan x\,dx=-\int\frac{du}{u}=-\ln|u|+C=-\ln|\cos x|+C=\ln|\sec x|+C.$

Answer: $-\ln|\cos x|+C=\ln|\sec x|+C$

Worked Example

Evaluate $\displaystyle\int (\ln x)^2\cdot\frac{1}{x}\,dx$.

Solution

Put $u=\ln x$, $du=\dfrac{1}{x}\,dx$. Then $\displaystyle\int u^2\,du=\frac{u^3}{3}+C=\frac{(\ln x)^3}{3}+C.$

Answer: $\dfrac{(\ln x)^3}{3}+C$

Worked Example

Evaluate $\displaystyle\int \sin^2 x\,dx$.

Solution

Power-reduce: $\sin^2x=\dfrac{1-\cos2x}{2}$. So $\displaystyle\int\sin^2x\,dx=\frac12\int(1-\cos2x)\,dx=\frac12\left(x-\frac{\sin2x}{2}\right)+C=\frac{x}{2}-\frac{\sin2x}{4}+C.$

Answer: $\dfrac{x}{2}-\dfrac{\sin2x}{4}+C$

Worked Example

Evaluate $\displaystyle\int \sin 3x\,\cos 2x\,dx$.

Solution

Product-to-sum: $2\sin3x\cos2x=\sin5x+\sin x$, so $\sin3x\cos2x=\dfrac12(\sin5x+\sin x)$. Then $\displaystyle\int=\frac12\left(-\frac{\cos5x}{5}-\cos x\right)+C=-\frac{\cos5x}{10}-\frac{\cos x}{2}+C.$

Answer: $-\dfrac{\cos5x}{10}-\dfrac{\cos x}{2}+C$

Worked Example

Evaluate $\displaystyle\int \cos^3 x\,dx$.

Solution

Odd power: keep one cosine, convert the rest. $\cos^3x=\cos x(1-\sin^2x)$. Put $u=\sin x$, $du=\cos x\,dx$: $\displaystyle\int(1-u^2)\,du=u-\frac{u^3}{3}+C=\sin x-\frac{\sin^3x}{3}+C.$

Answer: $\sin x-\dfrac{\sin^3x}{3}+C$

Worked Example

Evaluate $\displaystyle\int \frac{dx}{x^2+9}$.

Solution

Standard form with $a=3$: $\displaystyle\int\frac{dx}{x^2+a^2}=\frac1a\tan^{-1}\frac{x}{a}+C$. Hence $=\dfrac13\tan^{-1}\dfrac{x}{3}+C.$

Answer: $\dfrac13\tan^{-1}\dfrac{x}{3}+C$

Worked Example

Evaluate $\displaystyle\int \frac{dx}{\sqrt{9-4x^2}}$.

Solution

Write $9-4x^2=4\left(\tfrac94-x^2\right)$, so $\sqrt{9-4x^2}=2\sqrt{\left(\tfrac32\right)^2-x^2}$. Then $\displaystyle\int\frac{dx}{2\sqrt{(3/2)^2-x^2}}=\frac12\sin^{-1}\frac{x}{3/2}+C=\frac12\sin^{-1}\frac{2x}{3}+C.$

Answer: $\dfrac12\sin^{-1}\dfrac{2x}{3}+C$

Worked Example

Evaluate $\displaystyle\int \frac{dx}{x^2+4x+13}$.

Solution

Complete the square: $x^2+4x+13=(x+2)^2+9=(x+2)^2+3^2$. With $t=x+2$, $\displaystyle\int\frac{dt}{t^2+3^2}=\frac13\tan^{-1}\frac{t}{3}+C=\frac13\tan^{-1}\frac{x+2}{3}+C.$

Answer: $\dfrac13\tan^{-1}\dfrac{x+2}{3}+C$

Worked Example

Evaluate $\displaystyle\int \frac{2x+3}{x^2+2x+5}\,dx$.

Solution

The denominator's derivative is $2x+2$. Write $2x+3=(2x+2)+1$. Then the integral is $\displaystyle\int\frac{2x+2}{x^2+2x+5}\,dx+\int\frac{dx}{(x+1)^2+4}=\ln(x^2+2x+5)+\frac12\tan^{-1}\frac{x+1}{2}+C.$

Answer: $\ln(x^2+2x+5)+\dfrac12\tan^{-1}\dfrac{x+1}{2}+C$

Worked Example

Evaluate $\displaystyle\int \frac{x}{(x-1)(x-2)}\,dx$.

Solution

Set $\dfrac{x}{(x-1)(x-2)}=\dfrac{A}{x-1}+\dfrac{B}{x-2}$. Putting $x=1$ gives $A=-1$; putting $x=2$ gives $B=2$. So $\displaystyle\int=-\ln|x-1|+2\ln|x-2|+C.$

Answer: $-\ln|x-1|+2\ln|x-2|+C$

Worked Example

Evaluate $\displaystyle\int \frac{1}{(x-1)^2(x+2)}\,dx$.

Solution

Decompose $\dfrac{1}{(x-1)^2(x+2)}=\dfrac{A}{x-1}+\dfrac{B}{(x-1)^2}+\dfrac{C}{x+2}$. Solving gives $A=-\tfrac19,\ B=\tfrac13,\ C=\tfrac19$. Then $\displaystyle\int=-\frac19\ln|x-1|-\frac{1}{3(x-1)}+\frac19\ln|x+2|+C.$

Answer: $-\dfrac19\ln|x-1|-\dfrac{1}{3(x-1)}+\dfrac19\ln|x+2|+C$

Worked Example

Evaluate $\displaystyle\int \ln x\,dx$.

Solution

By parts with $u=\ln x,\ dv=dx$, so $du=\dfrac1x\,dx,\ v=x$. $\displaystyle\int\ln x\,dx=x\ln x-\int x\cdot\frac1x\,dx=x\ln x-x+C.$

Answer: $x\ln x-x+C$

Worked Example

Evaluate $\displaystyle\int x^2\,e^{x}\,dx$.

Solution

Apply parts twice. First $u=x^2,\ dv=e^x\,dx$: $\displaystyle\int x^2e^x\,dx=x^2e^x-2\int xe^x\,dx$. Using $\int xe^x\,dx=e^x(x-1)$ from Example 3, $=x^2e^x-2e^x(x-1)+C=e^x(x^2-2x+2)+C.$

Answer: $e^x(x^2-2x+2)+C$

Worked Example

Evaluate $\displaystyle\int e^{x}\left(\frac{1}{x}+\ln x\right)dx$.

Solution

This is $e^x[f(x)+f'(x)]$ with $f(x)=\ln x$ and $f'(x)=\dfrac1x$. By the special result $\int e^x[f+f']\,dx=e^xf(x)+C$, the answer is $e^x\ln x+C.$

Answer: $e^x\ln x+C$

Worked Example

Evaluate $\displaystyle\int \sqrt{4-x^2}\,dx$.

Solution

Use $\int\sqrt{a^2-x^2}\,dx=\dfrac{x}{2}\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac{x}{a}+C$ with $a=2$. Hence $=\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}+C.$

Answer: $\dfrac{x}{2}\sqrt{4-x^2}+2\sin^{-1}\dfrac{x}{2}+C$

Worked Example

Evaluate $\displaystyle\int \sqrt{x^2+16}\,dx$.

Solution

Use $\int\sqrt{x^2+a^2}\,dx=\dfrac{x}{2}\sqrt{x^2+a^2}+\dfrac{a^2}{2}\ln\left|x+\sqrt{x^2+a^2}\right|+C$ with $a=4$. Hence $=\dfrac{x}{2}\sqrt{x^2+16}+8\ln\left|x+\sqrt{x^2+16}\right|+C.$

Answer: $\dfrac{x}{2}\sqrt{x^2+16}+8\ln\left|x+\sqrt{x^2+16}\right|+C$

Worked Example

Evaluate $\displaystyle\int \frac{dx}{\sqrt{x^2-25}}$.

Solution

Standard irrational form with $a=5$: $\displaystyle\int\frac{dx}{\sqrt{x^2-a^2}}=\ln\left|x+\sqrt{x^2-a^2}\right|+C$. Thus $=\ln\left|x+\sqrt{x^2-25}\right|+C.$

Answer: $\ln\left|x+\sqrt{x^2-25}\right|+C$

Key Points

$\int f\,dx=F+C$ where $F'=f$; integration is linear and the constant $C$ must never be dropped.
Memorise the standard table: powers, $\tfrac1x\to\ln|x|$, exponentials, all six trig integrals, and the three inverse-trig results.
Substitution $u=g(x)$ targets "function and its derivative"; special cases $\int\tfrac{f'}{f}=\ln|f|$ and $\int f^n f'=\tfrac{f^{n+1}}{n+1}$.
Trig powers/products need an identity first: power-reduction $\sin^2x,\cos^2x$ for even powers, product-to-sum for mixed angles, peel-one-factor for odd powers.
Special forms: $\int\tfrac{dx}{x^2\pm a^2}$, $\int\tfrac{dx}{a^2-x^2}$, $\int\tfrac{dx}{\sqrt{a^2-x^2}}$ and $\int\tfrac{dx}{\sqrt{x^2\pm a^2}}$; complete the square for any quadratic denominator.
Partial fractions split a PROPER rational function — distinct linear, repeated linear $\tfrac{A}{x-a}+\tfrac{B}{(x-a)^2}$, and quadratic $\tfrac{Ax+B}{x^2+bx+c}$; divide first if improper.
By parts $\int u\,dv=uv-\int v\,du$ with $u$ chosen by ILATE; integrate lone $\ln x$ / $\tan^{-1}x$ with $dv=dx$; remember $\int e^x[f+f']\,dx=e^xf(x)+C$.
The two root integrals $\int\sqrt{a^2-x^2}\,dx$ and $\int\sqrt{x^2\pm a^2}\,dx$ are standard; verify any answer by differentiating it back.

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.$\displaystyle\int x^3\,dx=$

Explanation: $\int x^n dx=\tfrac{x^{n+1}}{n+1}+C$ with $n=3$.

Q2.$\displaystyle\int \tfrac1x\,dx=$

Explanation: The standard log integral.

Q3.For $\int xe^x dx$, by ILATE the choice $u=$

Explanation: Algebraic $x$ precedes Exponential, so $u=x$.

Q4.$\displaystyle\int 2x\,e^{x^2}dx=$

Explanation: Substitute $u=x^2$.

Practice more MCQs → 📝 Assignment · 35 marks